CHE230 Environmental Chemistry 2010
Tutorial Problem 3A Solutions
Consider a groundwater sample at 25 0oC in equilibrium with limestone and air at 1
bar containing 370 ppmv of CO2(g).
ΔHfo at 25oC (kJ/mol)
Sfo at 25oC (J/mol-K)
CaCO3(s)
-1206.9
92.9
Ca2+(aq)
-542.8
-53.1
HCO3- (aq)
-692
91.2
CO2(g)
-393.5
213.6
H2O(l)
-285.8
69.9
CO32- (aq)
-677.1
-56.9
0
0
H+(aq)
o
o
(assume ΔH and ΔS are independent of temperature over this temperature range)
a) Calculate the solubility product for CaCO3(s) at 25oC.
CaCO3 <=> Ca2+ + CO32-
Delta H= Delta Hf products – Delta Hf reactants = -542.8-677.1+1206.9= -13 kJ/mol
Delta S= Sf products – Sf reactants = -56.9-53.1-92.9= -202.9 J/mol-K
Delta G= DeltaH- T * DeltaS/1000 J/kJ
At 298K,
Delta G= 47.4642 kJ/mol
Therefore solving for solubility, Ksp=exp (-Delta G*1000/(8.324*T)) = 4.8*10-9
b) Calculate the equilibrium constant at 60oC for the reaction
CO2(g) + H2O  H+ (aq) + HCO3- (aq)
Since ΔHo and ΔSo are independent of temperature,
Delta H= Delta Hf products – Delta Hf reactants = -12.7 kJ/mol
Delta S= Sf products – Sf reactants = -192.3 J/mol-K
Delta G= DeltaH- T * DeltaS/1000 J/kJ
At 333K,
Delta G= 51.3359 kJ/mol
-9
K = exp (-Delta G*1000/(8.324*T)) = 8.9*10
c) Calculate the total alkalinity of the water if it is at 25 oC.
CaCO3 +CO2(g) + H20<=> Ca2+ + 2HCO3-
Delta H= Delta Hf products – Delta Hf reactants = -40.6 kJ/mol
Delta S= Sf products – Sf reactants = -247.1 J/mol-K
Delta G= DeltaH- T * DeltaS/1000 J/kJ
At 298K,
Delta G= 33.0358 kJ/mol
Therefore solving for solubility, K=exp (-Delta G*1000/(8.324*T)) = 1.6*10-6
Assume that Ca2+ = X and HCO3- = 2X
-4
CO2 (g) is given (370 ppmv or 3.70*10 bar)
2
-4
Therefore K = X * (2*X) / 3.70*10
-4
X= [Ca2+]25C = 5.7*10 M
-3
[HCO3-] = 1.1*10
-9
-4
We can find [CO32-] from a, since Ksp = 4.8*10 = [Ca2+]*[CO32-] = 5.7*10 * [CO32-]]
-6
Therefore, [CO32-] = 8.4*10
-3
Alkalinity = 2*[CO32-] + [HCO3-] = 1.2*10
d) 50 L of this groundwater water is heated to 60oC in a domestic water heater
causing the calcium to precipitate out as CaCO3(s):
Ca2+(aq) + 2 HCO3- (aq) <--------> CaCO3(s) + CO2(g) + H2O(l)
Estimate the mass of scale CaCO3(s) that is formed assuming that the CO2(g) in the
heater remains at 370 ppmv.
CaCO3 +CO2(g) + H20<=> Ca2+ + 2HCO3-
Delta H= Delta Hf products – Delta Hf reactants = -40.6 kJ/mol
Delta S= Sf products – Sf reactants = -247.1 J/mol-K
Delta G= DeltaH- T * DeltaS/1000 J/kJ
At 333K,
Delta G= 41.6843 kJ/mol
-7
K = exp (-Delta G*1000/(8.324*T)) = 2.9*10
Assume that Ca2+ = X and HCO3- = 2X
-4
CO2 (g) is given (370 ppmv or 3.70*10 bar)
2
-4
Therefore K = X * (2*X) / 3.70*10
-4
X= [Ca2+]60C = 3.2*10 M
Mass of CaCO3 precipitated = V * ([Ca2+]25C - [Ca2+]60C) * MW CaCO3
MW CaCO3 = 40+12+3*16
V=50L
[Ca2+]25C has been solved for in c
Therefore, mass of CaCO3 precipitated = 1.241 g
CHE230 Environmental Chemistry 2010
Tutorial Problem 3B Solutions
Consider a groundwater sample at 25 0oC in equilibrium with limestone and air at 1
bar containing 370 ppmv of CO2(g).
a) Calculate the solubility product for CaCO3(s) at 50oC.
CaCO3 <=> Ca2+ + CO32-
Delta H= Delta Hf products – Delta Hf reactants = -542.8-677.1+1206.9= -13 kJ/mol
Delta S= Sf products – Sf reactants = -56.9-53.1-92.9= -202.9 J/mol-K
Delta G= DeltaH- T * DeltaS/1000 J/kJ
At 323K,
Delta G= 52.5367 kJ/mol
Therefore solving for solubility, Ksp=exp (-Delta G*1000/(8.324*T)) = 3.2*10-9
b) Calculate the equilibrium constant at 25oC for the reaction
CO2(g) + H2O  H+ (aq) + HCO3- (aq)
Since ΔHo and ΔSo are independent of temperature,
Delta H= Delta Hf products – Delta Hf reactants = -12.7 kJ/mol
Delta S= Sf products – Sf reactants = -192.3 J/mol-K
Delta G= DeltaH- T * DeltaS/1000 J/kJ
At 298K,
Delta G= 44.6054 kJ/mol
-8
K = exp (-Delta G*1000/(8.324*T)) = 1.5*10
c) Calculate the total alkalinity of the water if it is at 25 oC.
CaCO3 +CO2(g) + H20<=> Ca2+ + 2HCO3-
Delta H= Delta Hf products – Delta Hf reactants = -40.6 kJ/mol
Delta S= Sf products – Sf reactants = -247.1 J/mol-K
Delta G= DeltaH- T * DeltaS/1000 J/kJ
At 298K,
Delta G= 33.0358 kJ/mol
Therefore solving for solubility, K=exp (-Delta G*1000/(8.324*T)) = 1.6*10-6
Assume that Ca2+ = X and HCO3- = 2X
-4
CO2 (g) is given (370 ppmv or 3.70*10 bar)
2
-4
Therefore K = X * (2*X) / 3.70*10
-4
X= [Ca2+]25C = 5.7*10 M
-3
[HCO3-] = 1.1*10
-9
-4
We can find [CO32-] from a, since Ksp = 4.8*10 = [Ca2+]*[CO32-] = 5.7*10 * [CO32-]]
-6
Therefore, [CO32-] = 8.4*10
-3
Alkalinity = 2*[CO32-] + [HCO3-] = 1.2*10
d) 50 L of this groundwater water is heated to 50oC in a domestic water heater
causing the calcium to precipitate out as CaCO3(s):
Ca2+(aq) + 2 HCO3- (aq) <--------> CaCO3(s) + CO2(g) + H2O(l)
Estimate the mass of scale CaCO3(s) that is formed assuming that the CO2(g) in the
heater remains at 370 ppmv.
CaCO3 +CO2(g) + H20<=> Ca2+ + 2HCO3-
Delta H= Delta Hf products – Delta Hf reactants = -40.6 kJ/mol
Delta S= Sf products – Sf reactants = -247.1 J/mol-K
Delta G= DeltaH- T * DeltaS/1000 J/kJ
At 323K,
Delta G= 39.2133 kJ/mol
-7
K = exp (-Delta G*1000/(8.324*T)) = 4.6*10
Assume that Ca2+ = X and HCO3- = 2X
-4
CO2 (g) is given (370 ppmv or 3.70*10 bar)
2
-4
Therefore K = X * (2*X) / 3.70*10
-4
X= [Ca2+]50C = 3.8*10 M
Mass of CaCO3 precipitated = V * ([Ca2+]25C - [Ca2+]50C) * MW CaCO3
MW CaCO3 = 40+12+3*16
V=50L
[Ca2+]25C has been solved for in c
Therefore, mass of CaCO3 precipitated = 0.979 g